# Solving Leetcode Interviews in Seconds with AI: Pow(x, n)


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "50" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> Implement pow(x, n), which calculates x raised to the power n (i.e., xn).   Example 1:  Input: x = 2.00000, n = 10 Output: 1024.00000  Example 2:  Input: x = 2.10000, n = 3 Output: 9.26100  Example 3:  Input: x = 2.00000, n = -2 Output: 0.25000 Explanation: 2-2 = 1/22 = 1/4 = 0.25    Constraints:  -100.0 < x < 100.0 -231 <= n <= 231-1 n is an integer. Either x is not zero or n > 0. -104 <= xn <= 104  

	# Explanation
	Here's a breakdown of the solution and the Python code:

*   **High-Level Approach:**

    *   **Handle Negative Exponents:** If `n` is negative, invert `x` and make `n` positive.
    *   **Binary Exponentiation:**  Use the property that x<sup>n</sup> = x<sup>n/2</sup> * x<sup>n/2</sup> if `n` is even, and x<sup>n</sup> = x * x<sup>n-1</sup> if `n` is odd. Recursively apply this to calculate the power efficiently.
    *   **Base Case:** If `n` is 0, return 1.

*   **Complexity:**
    *   Runtime Complexity: O(log n) due to the halving of the exponent in each recursive step. Storage Complexity: O(log n) due to the recursive call stack.

	
	# Code
	```python
	def myPow(x: float, n: int) -> float:
    """
    Calculates x raised to the power n (x^n).
    """

    if n == 0:
        return 1.0

    if n < 0:
        x = 1 / x
        n = -n

    def power(x, n):
        if n == 0:
            return 1.0
        if n % 2 == 0:
            half_power = power(x, n // 2)
            return half_power * half_power
        else:
            return x * power(x, n - 1)

    return power(x, n)
	```